CHAPTER 11

THREE-PHASE SYNCHRONOUS MACHINES

11.1 Introduction In the majority of cases, the rotor has the d.c. winding and the stator the a.c. winding. An alternator with a It was shown in an earlier chapter that an alternator rotating a.c. winding and a stationary d.c. winding, while driven at a constant speed produces an alternating voltage suitable for smaller outputs, is not satisfactory for the at a fixed frequency dependent on the number of poles in larger outputs required at power stations. With these the machine. A machine designed to be connected to the machines the output can be in megawatts; a value too supply and run at synchronous speed is referred to as a large to be handled with brushes and slip rings. Because synchronous machine. The description applies to both the terminal voltages range up to 33 kV, the only motors and generators. A synchronous condenser is a satisfactory construction is to have the a.c. windings special application of a synchronous motor. stationary and to supply the rotor with d.c. While the synchronous motor has only one generally This arrangement has the following advantages: used name, the synchronous generator is on occasion referred to as an alternator or as an a.c. generator. The l. extra winding space for the a.c. windings; term alternator has been used in previous chapters and 2. easier to insulate for higher voltages; will be used in this chapter but it should be remembered 3. simple, strong rotor construction; that other terms are in use. 4. lower voltages and currents in the rotating windings; In general, the principles of construction and 5. the high current \Vindings have solid connections to the operation are similar for both alternators and generators, "outside" circuit; just as there were basic similarities between d.c. motors 6. better suited to the higher speeds (and smaller number and generators. of poles) of turbine drives. While alternators were once seldom seen outside power houses and whole communities were supplied from 11.2.1 Stator a central source, there is now an expanding market for The stator of the three-phase synchronous machine smaller sized alternators suitable for the provision of consists of a slotted laminated core into which the stator power for portable tools. Today, with the growth in winding is fitted. The stator winding consists of three computer control, there is a further need for standby separate windings physically displaced from each other by generating plant to ensure a continuity of supply to 120° E. Each phase winding has a number of coils prevent a loss of data from computer memories. So much connected in series to form a definite number of magnetic information is being stored in computers today that even poles. A four-pole machine, for example, has four groups brief interruptions to the power supplies can have serious of coils per phase or four "pole-phase groups". The ends consequences on the accuracy and extent of information of the three phase windings are connected in either star or stored. delta to the external circuit. Details of phase windings for a three-phase machine were shown in Chapter l 0 to consist of three identical 11.2 Alternators windings symmetrically distributed around the stator. The three-phase synchronous machine has two main windings: 11.2.2 Rotor 1. a three-phase a.c. winding; The alternator rotor can be of two types-low speed and 2. another winding carrying d.c. high speed. 209 210 ELECTRICAL PRINCIPLES FOR THE ELECTRICAL TR 11.2.3 Prime movers Low speed Most diesel engines used as prime movers for alternators operate within the speed range of 50 r /min and this necessitates the use of rotors witl pairs of poles. Hydroelectric turbines have water-driven in which operate at low speeds, consequently they all rotors with. many poles. While the diesel-driven alt usually has its shaft in the horizontal plat hydroelectric unit has its shaft in the vertical plar method of construction means that special thrust t have to be fitted to take the end thrust of the 1 component.

High speed Turbine prime movers, whether steam or gas, efficiently at speeds in the vicinity of 3000 r / n alternator driven by a turbine and producing a fn of 50 Hz at 3000 r /min must consist of only two Fig. 11.1 Stator for a tour-pole 415 V three-phase 350 kVA In Chapter 6 the relationship between alternator frequency and the number of poles was shown tc DUNLITE GENERATING SET MANUFACTURERS ~ Low speed (salient pole) ~ This type usually consists ofa "spider" similar to that used By transposition in d.c. machines, on which are bolted the field poles and 120/ the field coils (see Fig. l l .2(a)). Physical constraints limit n=-- the use of this type of rotor to low-speed machines. p where n = r/min High speed (cylindrical) f = frequency in hertz The cylindrical rotor was developed to meet the needs of p = number of poles higher-speed prime movers. To counteract centrifugal For a large-diameter rotor of twenty-four 1 forces its diameter must be small in comparison to its 50 Hz length (see Fig. l l.2(b)). 120 50 n = X = 250 r/min 24 For a turbine-type rotor of two poles at 50 f 120 50 n = x = 3000 r/min 2

(b)

Fig. 11.2 Main types of alternator rotors: (a) low speed-salient pole, (b) high speed-cylindrical THREE-PHASE SYNCHRONOUS MACHINES 211

Example 11.1 The diameter of the rotor must be small enough to At what speed would the governor of a twelve-pole diesel­ keep the surface speed down to a safe value, so for large driven alternator have to be set to enable a frequency of capacities the length of the machine must be considerable. 60 Hz to be generated? This long axial length causes difficulty in cooling the central portion of the core, because the heat generated n = 120{ cannot be conducted away quickly enough to limit the p temperature rise in the core to a value that will protect the 120 x 60 windings and the insulation. = 600 r/min 12 These considerations gave rise to the necessity for completely enclosing the alternator, and allowing the use An alternator in this speed range will have a large of forced ventilation to carry away the heat produced. diameter and have a comparatively short axial length. Where cooling air is used, it must be filtered to keep it With turbines, the extra expense and auxiliary machinery clean and sometimes washed by passing it through a spray needed restricts their use to larger sizes. Higher outputs chamber to prevent a build-up of dust within the machine. mean that the length of the alternator must be increased Washing the air has the added advantage of cooling it, and the increase in length causes complications in cooling. and so further reducing the temperature of the alternator and allowing the rating of the machine to be increased. 11.2.4 Alternator cooling To increase alternator ratings still more, hydrogen gas Low speed is used instead of air because ofits greater ability to absorb With engine-driven or hydroelectric alternators, there is heat. The machine is completely enclosed and the no great difficulty in providing adequate ventilation hydrogen is blown through the alternator and then ~ because of the characteristically large diameter and short through a heat-exchanger before being cycled through the (" ~ axial length. In addition to the large surface area available alternator again. The total exclusion of air from a fully J­ for direct radiation of heat, there is a fanning action due to sealed machine is necessary to prevent an explosive '­ the rotation of the fields; an action which can be increased mixture from forming. 3 These cooling methods require considerable powe 1o::: by the addition of fan blades if necessary. 0 When the axial length is short, the heat developed in and auxiliary equipment, so the output from th \ the imbedded windings is quickly conducted to the ends, alternator must be increased an appreciable amount fo \;, where the fanning action can dissipate it. As the machine the method to be economically feasible. Accordingly. it is '. "£,. size becomes larger, it is often necessary to provide only used on very high-capacity machines. ~ ~ _,, ventilation ducts within the core to provide paths through which cooling air can flow. 11.2.5 Excitation ~ The usual method for d.c. excitation of the rotor High speed \Vindings is for each machine to have its own d.c. The provision of adequate cooling facilities is a problem in generator called an "exciter" (refer to Fig. 11.3). The high-speed machines of large capacity if the operating exciter can be belt-driven or geared down from the temperature of the windings is to be kept within safe synchronous machine. but the usual practice is for the limits. The surface area available for cooling in a high­ exciter to be directly coupled to the rotor shaft. speed machine is less than that in a low-speed machine of The exciter armature rotates within the influence of the same capacity. the exciter field, causing a d.c. voltage to be generated in

,------Exciter --1 1------1Alternator I I I I I t Stator Brush windings I I gear Exciter I and Rotor I field slip field I I rings I I I I I I I L ______J L ______J

Fig. 11.3 Basic alternator circuit 212 ELECTRICAL PRINCIPLES FOR THE ELECTRICAL TR the armature. The exciter output is fed into the field where V, = generated voltage per phase (r.m.s.) windings of the synchronous machine. By adjusting the IP = flux per pole in webers rheostat in the exciter field circuit, the strength of the f = frequency in hertz magnetic field in the rotor can be varied. N = number of turns per phase With very large alternators the d.c. excitation kd = a constant, dependent on winding di requirements are substantial. This means that the d.c. ti on generators have to be large also; so large that they may not kp = a constant, dependent on coil pitch be able to self-excite. Because of this, the generator may need an exciter of its own-one that is able to self-excite Example 11.2 and provide power for the field of the main d.c. generator Calculate the line voltage of a 50 Hz star-cor which in turn supplies the rotor field of the alternator. alternator given the following details: Some alternators use "brushless" excitation in which the exciter armature has been replaced by a three-phase IP = 0.67 Wb/pole winding which rotates within the influence of a d.c. Kd = 0.85 magnetic field, causing a three-phase voltage to be KP = 0.98 generated in the exciter. This three-phase exciter output is N = 36 turns/phase fed through a full-wave bridge rectifier, mounted on the V, = 4.44 IPJN KdKp end of the exciter and converted to d.c. = 4.44 x 0.67 x 50 x 36 x 0.85 The resulting d.c. is in turn fed into the rotor windings = 4460 v of the synchronous machine. By varying the current through the exciter field, the rotor field is varied and so Then V, = yJ x Vp governs the value of the generated voltage (see Fig. 11.4). = I. 732 x 4460 = 7725 v 11.2.6 Generated voltage The value of the generated a.c. voltage depends on the 11.2. 7 Effect of load on alternator voltage strength of the rotor flux and the speed at which it cuts the An alternator can be considered to consist c windings. Because the speed must be constant (and is components in series: linked to the frequency required), the sole remaining I. an a.c. generating source; factor determining the value of the generated voltage is the 2. a resistor-representing iron and copper losse~ strength of the rotor flux. 3. an inductor-representing the inductance For an alternator the generated voltage is found from: windings and magnetic leakage. Any load placed on the alternator must be assum v, = 4.44 IPJNkdkp I in series with these components as shown in Fjgu

Three-phase bridge rectifier

+ Three-phase exciter windings Rotor Stator field windings Exciter field

1. Rotating components .1 Fig. 11.4 Brushless excitation THREE-PHASE SYNCHRONOUS MACHINES 213 generated voltage v,. For a load with a lagging power ,---- l factor, however, the magnetic effect of the stator currents I I opposes that of the rotor, resulting in a weakened rotor I I field and reducing the output voltage further than did the resistive load (see Fig. l l .6(b)). As before, IR is in phase I I with the load current!. /Xis at 90°E to JR so placing /Z at I I I R I Alternator Load z a different angle to the previous case. In a similar manner, I V, is equal to the phasor sum of the output voltage and /Z. I For a load with a leading power factor, the flux caused by I I the stator currents assists that of the rotor resulting in an I L I increased output voltage (see Fig. l l.6(c)). The I I __ J characteristics of the three types of loads are shown in L _ Figure 11.7.

Fig. 11.5 Equivalent circuit of an alternator 11.2.8 Voltage regulation The series impedance of the resistance and inductance An alternator is required to give a prescribed terminal provides a drop in voltage before the generated voltage voltage at full load. The difference in output between no can reach the connected load. Additionally the load load and full load is a measure of its voltage regulation. current in the a.c. windings produces an armature The difference is compared to the full-load value in a reaction which also affects the output voltage. similar manner to that for d.c. machines. With a unity power factor load the armature reaction . [VNL- VFL ] merely distorts the main field and the effect on voltage is Voltage regulat10n = VFL X 100 % minimal, the voltage drop in the main being due to the series impedance. Figure l I.6(a) shows that the resistive voltage drop IR is in phase with the load current I and the Example 11.3 voltage drop due to the reactance IX is at 90°E to the IR A three-phase star-connected alternator has an output drop. These two values combine to form a voltage drop IZ voltage of 3300 V at full load with unity power factor. due to the impedance of the alternator windings. The When the load is removed and the excitation is unchanged phasor sum of the output voltage and /Z gives the the voltage rises to 3350 V. Find the percentage regulation. v,_ Change in voltage = 3350 - 3300 = 50 V . 50 x 100 Regulal!on = 3300 IZ : IX = I.5% at unity power factor. v. Note The regulation must also be referred to the load IR power factor because these figures at any other (a) Unity power factor power factor would be different. v,

IZ_./.:

V _.·IX -E---.--7· ·.: Leading power JR Output voltage E=::::::::==:::~~;~-=~; factor (b) Lagging power factor r Unity power factor

Lagging power factor

(c) Leading power factor Load current

Fig. 11.6 Phasors for various power factor loads on an Fig. 11.7 Effect of power factor on output voltage of an alternator alternator 214 ELECTRICAL PRINCIPLES FOR THE ELECTRICAL TRI

11.2.9 Alternator ratings Purchase price An alternator is rated according to three basic factors: The overall cost for smaller units may be lower, I. frequency; terms of cost per kVA they are more expensb 2. voltage; operate at lower efficiencies. As the size of tt 3. current. increases, the cost per kV A reduces while the op1 efficiency increases. The first fixes the speed at which the alternator must be driven; the second states the designed output voltage; and the third is the full-load current output. The last two Type of prime mover factors help establish the volt-ampere rating, usually The economy of the prime mover in terms of effi expressed in kV A. has a bearing on its selection. This in turn is affec The power factor of any load placed on the alternator the type of service it will encounter. For exarr is beyond the control of the manufacturer and because it steam turbine has a good economy throughout its could vary considerably, the alternator rating cannot be load range. However, it is expensive, large, and n given in kilowatts. long time to get the unit on load from cold. An i1 combustion engine has poor efficiency at light load, Example 11.4 much cheaper to buy initially. For some loads it is c to buy several smaller alternators than one larg' Find the power loading in kilowatts of a three-phase, 415 Problems of paralleling the units then have V, 50 Hz alternator rated at 150 kV A at 0.8 power factor, considered (see sect. 11.4). The cost and availab if the load has a power factor of: fuel must always be a consideration. While disti (a) 0.8; more expensive initially, as is the diesel engine its1 (b) 0.6. fuel cost per hour is less while maintenance costs higher than those for a petrol engine. The petrol er Machine is rated at 150 kVA and 0.8 power factor, cheaper to buy, the fuel is readily available, and the then at this load: suited to smaller units used purely for portable power output = 150 x 0.8 = 120 kW supplies on intermittent duties. In the long term th( engine runs better on full loads than the petrol engir At 0.6 power factor, petrol engine is more tolerant of dirty fuel than th< engine and does not need specialised skil power output = 150 x 0.6 = 90 kW maintenance purposes.

In both cases, the current flowing will be the full-load Starting methods current value, which should not be exceeded because of These are governed by the intended use of the geni cooling problems within the windings. unit. The quicker the changeover to auxiliary pm At 0.8 power factor, more expensive is the starting method. The ct P = VJ VIA. method involves merely starting the unit manually 1 i.e. 120 000 = ;j3 x 415 x Ix 0.8 is realised that the main power supply has failed. f :. I= 120000 = 208 A V3 x 415 x 0.8 ~ This is the full-load current rating for each phase winding of this particular alternator and it applies irrespective of the load power factor or of the load power.

11.3 Emergency power supplies and portable alternators The factors affecting the buying and running of alternators can be many and varied. They range from buying a small portable unit at the best possible price to careful planning for the most suitable unit for a particular purpose. It is not enough to simply select an alternator with respect to the load it has to supply; the choice should Fig. 11.8 A self~contained portable power supply. 7 be affected by many other considerations. Some of these alternator rated at 6 kVA is driven by a pet engine. The size and weight of the unit is. factors are listed below and their order of importance is that it can be carried to any site where po· governed by the actual use intended for the alternator. required. DEPT OF A' THREE-PHASE SYNCHRONOUS MACHINES 215 expensive method involves the use of a changeover 11.4 Parallel operation of alternators contactor which drops out when the main supply fails. In turn this connects a starting motor to the engine and after -synchronising the alternator has got up to speed connects it to the load. Most commercial power stations are designed to have a At the top end of the scale is the so-called "no-break"set. number of alternators operating in parallel, supplying a The alternator with a heavy flywheel is run as a syn­ common load at constant voltage. Because alternator chronous motor, being separated from the prime-mover efficiency is maximum near its full-load capacity, it is by an electrically operated clutch. When a mains failure more economical to have each machine delivering its occurs the clutch is released connecting the alternator and approximate rated output. During the early hours of the the flywheel to the engine. The engine is quickly run up to morning, for example, when there is a light load, it may be speed and the alternator reverts to its intended purpose. necessary to have only one machine connected to the line, The changeover period can be short enough to ensure delivering its rated output. As the load varies during the continuity of operation of essential equipment. The 24-hour period, so the number of machines connected in method is very expensive with high operating costs. parallel is determined. Before a three-phase alternator can be connected in Load sizes and alternator capacities parallel with another three-phase supply, the followin2 Smaller generating plant is usually intended for standby conditions must be fulfilled: purposes for short periods. It usually has only one load 1. The output waveform of each supply must be identical. connected to it at a time, such as a portable tool or a small This is determined by the design features of the lighting load. With middle- and larger-sized alternators alternators. It is standard practice to generate a consideration has to be given to the possible connection of sinusoidal waveform supply. intermittent larger loads, such as the starting currents of 2. The phase sequence or' rotation of each supply must be motors. The unit then has to have the electrical capacity the same and this ensures that the e.m.fs of each supply and engine power to maintain both the output voltage and reach their maximum values in the same sequence; for frequency during these current surges to avoid inter­ example, R, W, B. The phase sequence is determined ruptions to other equipment connected to the same by the method of connection of the alternator phase supply. windings to the terminals of the machine. This check is Operation of alternators carried out during the commissioning process after the With the exception of some manually operated equip­ initial installation, or following a major maintenance ment, most operations today are beyond the control of the overhaul, and it is not necessary to do it each time the operator. Where some degree of manipulation is available machine is connected in parallel with others. there are two important factors that should always be 3. The alternator and supply voltages must be the same. considered-voltage and frequency. In most cases the 4. The alternator and supply voltages must also be in voltage is governed by automatic voltage regulators while phase. the frequency is controlled by the engine governor. The 5. The alternator and supply frequencies must be order of operation is to set the speed first, which in turn identical. sets the frequency, and then adjust the voltage of the unit. The last three conditions are explained in Chapter 10. To do this in the reverse order is to alter the voltage each The value of the voltages, their phase relationship and time the speed is altered. their frequencies can be adjusted by the operator. The

Fig.11.9 An alternative form of "no~break unit". The unit shown has a 1470 r.p.m. motor· driving an alternator at all times. The generated frequency is 49 Hz so the unit cannot be used to supply frequency sensitive equipment. The photograph shows the alternator, driving motor, flywheel, clutch and diesel engine. When a power failure occurs, the flywheel keeps the alternator rotating tor 7 seconds allowing the diesel engine to start. The electrically operated clutch then connects the engine to the alternator. The interruption of power to the load is in the order of 0.5 cycle. DEPT OF AVIATION 216 ELECTRICAL PRINCIPLES FOR THE ELECTRICAL TRi

d.c. d.c. Synchronising Incoming Alternator on load lamps,-- alternator

--~\\--

'------+~._-----Three-phase distributions

Fig. 11.10 "Three dark" lamp method for synchronising alternators

voltage of the incoming alternator is adjusted by varying The lamps flicker at a rate equal to the diffen 1111 the field excitation, and the frequency is determined by the frequency between that of the incoming alternator' speed of the prime mover. busbars leading to the distribution system. 1 To ensure that the alternator and supply voltages are alternator frequency approaches that of the busb• in phase with each other before connecting them in rate of flickering slows down; when the two freq1 1111 parallel to the load, some method of indicating the phase are equal, the flickering stops. When the lamps ' relationship is required. Smaller-sized alternators can be (dark), the connecting switch can be closed and t synchronised with lamps, but for larger machines a more machines will remain synchronised. When all the exact method is required. are dark, there is no potential across the lamps, ind that the two voltages are in phase with each other The disadvantage of this connection is that th< can be dark even with a "small" voltage across ther 11.4.1 Synchronising alternators with smaller alternators the two a.c. sources can sync! incandescent lamps themselves ifthe difference is not too great, but wit! "Three dark" method alternators the mechanical and electrical forces ere: Voltages for synchronising purposes can be checked by a phase displacement between the two sources ca1 connecting a voltmeter to each machine in turn, but this considerable damage. does not give any indication of polarities or phase relationships. Incandescent lamps can be used to indicate "Two bright, one dark" method this and the circuit is shown in Figure 11.10. The circuit for this method is shown in Figure 11. The voltage rating of the lamps needs to be twice the can be seen to be similar to that of the previous alternator phase voltage, and the simplest way to achieve except that the connections for two of the Ian this is to connect two lamps of equal wattage in series. The crossed. Again two lamps are in series and it is t lamps can be observed as three pairs oflamps or three can cover up three lamps, leaving only three visible (as be covered, leaving only three visible (as shown in the box by the dotted lines). in the diagram). To use this circuit it is essential to check thj If the alternator is properly connected, the three lamps rotation by the "three dark" method first. should all become bright and dim simultaneously. If they established that the phase rotation is correct i brighten and dim in sequence, it means that the phase lamps reconnected, it will be found that the lamps! rotation of the alternator is opposite to that on the and bright in sequence. By noting the order of bri! distribution system, so the phase rotation of the incoming it becomes a reference in determining whetl alternator must be reversed. incoming alternator is fast or slow. THREE-PHASE SYNCHRONOUS MACHINES 217

d.c. d.c. Synchronising Incoming Alternator on load lamps alternator ,--1 I I

L------~-+-----_.. Three-phase distribution system

Fig. 11.11 "Two bright, one dark" lamp method for synchronising alternators

Synchronism occurs when the lower lamp in Figure synchronism the pointer will remain stationary, but it 11.11 is dark and the other two are of equal brilliance. must be brought to an indicated position on the scale Then the switch can be safely closed. before the main switch of the incoming alternator is The significance of the correct lamp being dark lies in closed. the fact that it is connected between two similar phases. When these two phases are synchronised, the voltage difference between them is zero. This cannot apply to the 11.5 Hunting in alternators other lamps since they are connected across dissimilar The driving torque of a diesel engine is not uniform during phases. a revolution of the flywheel but varies according to ihe The "two bright, one dark" method gives greater positions of the pistons. Even with a heavy flywheel, the accuracy-both in determining the relative speeds and variation in torque can result in changes in the induced frequencies, as well as showing fairly accurately the voltage. These voltage pulses can cause circulating instant for synchronising. currents to flow between alternators in parallel, resulting

11.4.2 Synchronising alternators with a synchroscope A synchroscope is an instrument that indicates both phase relationships and relative speed for an incoming alternator. There are variations between manufacturers for the operating principles, but in general a synchroscope consists of a two-phase stator connected to the incoming alternator with the rotor wound with a polarising coil and connected to the supply source. Some models use rotating vanes with no actual electrical connection to the rotor. The synchroscope front panel is shown in Figure 11. 12 and the connections are shown in Figure 11.13. If there is any difference between the frequencies of the supply and the incoming alternator, a pointer attached to the rotor of the synchroscope will rotate at a speed Fig.11.12 Portable synchroscope contained in a polished proportional to this difference. Its direction of rotation wooden box. At the moment of synchronisation indicates whether the incoming machine is running fast or the pointer should be stationary over the vertical line. slow (i.e. above or below synchronous speed). At A. J. WILLIAMS 218 ELECTRICAL PRINCIPLES FOR THE ELECTRICAL TRI d.c. d.c.

Incoming Alternator on load alternator Synchroscope

'-+------4-+------Three-phase distributions

Fig. 11.13 Synchroscope connection for synchronising alternators

1!11· in mechanical oscillations. With turbines, the pulsing or polarity: one field is that of the rotating stator a hunting is usually due to fluctuations in the governor other that of the rotor. settings with changes in load. Remedies for hunting A synchronous motor has torque only at synch involve the use of heavy flywheels and special windings in speed, so special steps have to be taken to get the 11111 the pole faces. These windings are discussed in more detail up to speed and synchronised with the supply. Tl in the next section. magnetic fields are then rotating at the same spe1 lock in with each other. A later section in this c 11.6 Synchronous motors discusses starting methods for synchronous motors 11.6.1 Construction Stator 11.6.3 Effect of load on a synchronous mo The stator of a synchronous motor has a three-phase When a synchronous motor runs on no load, the I winding, as described in a previous chapter, and is of the positions of stator and rotor poles coincide as sh same type as that in an alternator. Figure l l.14(a). When this winding is energised with a.c. it produces a When a load is applied, the rotor must still cont magnetic flux, which rotates at a speed called the rotate at synchronous speed but due to the rel synchronous speed. It is the same speed at which the action of the load, the rotor pole lags behind th< synchronous motor would have to be driven to generate pole. Their relative positions are displaced by the: an a.c. voltage at line frequency. (called the "torque" or "load" angle), as shown in The speed can be derived from the same formula used l l.14(b). The greater the load applied, the lar: for alternators in section 11.2.3. torque angle. The magnetic coupling between each stator an Rotor pole distorts according to the load applied. If the I Although of similar construction to the alternator rotor, it the motor becomes excessive, the magnetic C< is usually made with salient poles. When excited with d.c. breaks and the rotor slows down until it stops. it produces alternate north and south magnetic poles, When the motor is rotating at synchronous which are attracted to those produced in the stator. with a fixed d.c. excitation in the rotor windings, tt flux cuts the stator windings, inducing a voltage 11.6.2 Operating principle phase winding and opposing the applied voltage A synchronous motor works on the principle of magnetic law). The phase relationship between this induced attraction between two magnetic fields of opposite and the applied voltage depends upon the THREE-PHASE SYNCHRONOUS MACHINES 219

Rotation of stator Torque angle I

Rotor \ \. ·~ I (a) No load (b) Loaded Fig. 11.14 Relative positions of stator and rotor magnetic fields in a synchronous motor positions of each stator and rotor pole, which in turn For a fixed rotor winding excitation, an increase in depend upon the load applied to the motor. load on a synchronous motor will therefore cause an Neglecting motor losses, on no load the torque angle is increase in current drawn from the supply .. with a poorer zero, and so the induced voltage V, and the applied power factor. voltage V are equal and opposite. The resultant voltage VR across the windings is zero, and so the current drawn from the supply is also zero. This is illustrated by the 11.6.4 Effect of varying field excitation phasors in Figure l l .15(a). If the load applied to a synchronous motor is constant, the When a light load is applied to the motor, the torque power input to the motor is also constant. angle a increases, and the induced voltage Vg in the stator When the rotor field excitation is varied, the induced windings is now ( 180 - a)0 E out of phase with the applied voltage in each stator winding is also altered. voltage V, as shown in Figure l l. l 5(b ). These two The phasor diagram in Figure l l.16(a) represents the voltages combine to produce an effective voltage v. conditions for a given load at unity power factor. The across the stator windings, which is sufficient to draw a power input per phase is Vl1. If the rotor field excitation is current I from the supply. Because of the relatively high decreased, the induced voltage Ve decreases, as shown in inductance of the stator windings, the line current /in each Figure l l. l 6(b ). This causes the line current h to lag the winding lags each resultant voltage v. by nearly 90°E. applied voltage Vby <1>2. Since the load, and so the power This causes the line current I to lag the applied voltage input, is constant, the power component of /2 must remain by cl>. the same as Ii in Figure l l.16(a). The line current Ii must As the load is increased, so the torque angle is increase to accommodate the lagging power factor. A increased. This causes an increase in the resultant voltage reduction in the d .c. field excitation therefore causes an VR across each stator winding, as seen in Figure l l.15(c). increase in line current, and a lagging power factor. Because of the increase in the value of VR• the line current If the d.c. excitation is increased, the induced voltage I increases, and the phase angle ¢ between the applied V,increases as shown in Figure l l.16(c). The line current voltage V and the line current I also increases. h will therefore lead the applied voltage V by ci>J, and will

v, v

(a) No load (b) Light load (c) Heavier load

Fig. 11.15 Effect of load on line current with constant excitation 220 ELECTRICAL PRINCIPLES FOR THE ELECTRICAL TR, Where large amounts of power are being dist1 and power factor correction is needed, specially d< synchronous motors are run without any load con1 Under these circumstances the overexcited synch motor is called a synchronous condenser. Voltage control v An important application is in the control of volt: I(a) Unity power factor transmission lines. Synchronous motors are inst::: suitable positions along the line and their exc adjusted as desired to cause them to draw lagi leading currents in order to raise or lower the \I When synchronous motors are installed unde1 conditions, there is a tendency to greater stab voltage on the transmission line. Low-speed drives I A synchronous motor has good efficiency and ] (b) Lagging power factor speeds its higher initial cost is adequately compens ,, the comparatively lower running cost. At low spe1 I v induction motor has a decreasing efficiency, wl vg -+-----:, synchronous motor retains its high efficiency. ',, 11.6.6 Hunting in synchronous motors A change in load on a synchronous motor causes a ' '- in the value of the torque angle (Fig. 11.14). In! ' the inertia of the rotor prevents an instant chang1 new conditions. with the result that the rotor shi V ihe point- of equilibrium and then has to correc (c) Leading power factor While the rotor and the rotating field in the stator rotating at a synchronous average speed, the ch Fig. 11.16 The effect of varying the d.c. excitation load on the rotor causes this periodic swing aro point of equilibrium. This surging or hunting ca also be greater than Ii in Figure l l.16(a) because the undesirable fluctuation in line current to the mot power component is the same, due to the load remaining The usual method for damping these surges i constant. An increase in d.c. excitation therefore causes damper winding, called an amortisseur wine an increase in line current and a leading power factor. consists of copper bars embedded in the pole fac< This characteristic of the synchronous motor, where rotor and shorted out at each end (Fig. 11. I" its power factor can be altered by varying its d.c. excitation, gives rise to its main application in industry­ /Dampe• power factor correction.

11.6.5 Applications of synchronous motors IT:. I Power factor correctiOn rQ .. o - - - If a synchronous motor has sufficient d.c. excitation to cause it to draw a leading current from the supply, the effect is one of power factor correction for other loads within an installation. - A motor running with a leading power factor is called overexcited, and is often designed to run as a synchronous --s motor driving a load and correcting overall power factor at the same time. The driven load selected is usually one in demand throughout the installation (e.g. air compressors, hydraulic systems or frequency changers for portable tools). . --- An added advantage can be an economical incentive loo o offered by supply authorities for ensuring a certain - - - minimum value power factor in an installation. For 11 •• ·1 example, the charge per kWh may be reduced if the power factor does not drop below 0.75 or some similar figure. Fig.11.17 Salient pole with amortisseur wind THREE-PHASE SYNCHRONOUS MACHINES 221 surging causes an induced voltage in the copper bars. This connected to the supply. It is an expensive method. results in a magnetic field being created and opposing the particularly if high starting torques are required. surging effect. Often the shorting-out bars are extended around the Induction motor starting rotor. resulting in a squirrel cage-type rotor winding A reduced line voltage is applied to the stator windings about the salient poles. While damping any tendency of and the d.c. winding on the rotor is short-circuited. With the rotor to hunt, they can also assist the motor in starting. the aid of the amortisseur winding, the complete machine behaves as an induction motor as it accelerates up to a 11.6. 7 Starting methods for synchronous motors speed slightly below synchronism. At an appropriate time Auxiliary motors the short is removed from the rotor winding. d.c. is Some synch'ronous motors are equipped with a special applied and the full line voltage applied to the stator motor designed for use during the starting period only. winding. Because the speed is only slightly less than The auxiliary motor runs the synchronous motor up to synchronous speed, the rotor field is able to lock in with speed, at which stage it is first synchronised and then the stator field and accelerate to synchronism.

Exercises 11.1 What advantages are there in using the 11.8 What is meant by the term "phase sequence" rotating d.c. field-type construction for when applied to three-phase synchronous synchronous machines? alternators? 11.2 What are the constructional differences 11.9 In what way does the principle of operation of between low-speed and high-speed alterna­ a synchronous motor differ from that of an tors? induction motor? 11.3 Explain why a low-speed synchronous 11.10 Why is a synchronous motor not self­ machine has a large salient pole-type rotor. starting? 11.4 What is the purpose of the "exciter"? 11.11 State two characteristics that are applicable only to a synchronous motor. 11.5 How many poles must a synchronous machine have to operate at 250 r/min and a 11.12 Explain how an increase in the load applied to frequency of 50 Hz? a synchronous motor affects the line current 11.6 How does the power factor of the load affect and power factor. the output voltage of an alternator? 11.13 How can the power factor of a synchronous 11. 7 State five conditions that must be satisfied motor be changed? before an alternator can be synchronised 11.14 What are some applications for synchronous with an existing supply. motors? CHAPTER12 THREE-PHASE INDUCTION MOTOR!

12.1 Introduction The stator core is held in the motor frame whi serves to carry the bearings holding the rotor, to The majority of a.c. motors used in industry are of the the coils and to provide a means whereby the wholl induction type. They are rugged and have a high degree of mounted (see Fig. 12.1 ). reliability. A three-phase induction n1otor consists of a The motor frame takes various forms, depen1 laminated stator with three identical windings placed the conditions under which the motor will aper; symmetrically in slots within it. The rotor is also open-type frame allo\vs free ventilation to take laminated, and usually has single-turn conductors placed drip-proof frame has a closed upper half, while a within its slots and short-circuited at the ends. To achieve ventilation through the lower half; a totally enclm special characteristics, conventional windings are some­ prevents the exchange of air between the inside times used instead. The motor derives its name from the outside of the frame. fact that the currents flowing in the rotor are induced and not drawn directly from the supply. 12.2.2 Motor enclosures 12.2 Construction The conditions governing the actual installatio induction motor are normally beyond the contrc 12.2.1 Stator motor manufacturer. As a result the motor is The laminated stator core is made up from sheet steel factured in various enclosures. A motor dr punchings with slots on the inner surface. The windings compressor for a refrigerated display cabir consist of three identical windings, laid out in the same example, may operate under such clean and d fashion as the alternator and synchronous motor. In ditions that the motor enclosure need only pr motors of higher power ratings the stator slots are of the mounting for the bearings and a means for fi:;. open type to allow the insertion of pre-shaped and motor in a horizontal plane. At the same time insulated coils, but in smaller sizes the slots are partially closure provides mechanical protection against ac closed to reduce the air gap as much as possible. spillage and enables cooling air to circulate freely· the motor windings. Compare this situation with a water turbin used for irrigation purposes. In most cases the t mounted vertically at the bore head and is g protection from the weather. The motor neeC totally enclosed to prevent the entry of water and is by means of heat transfer through the motor 1 The air sealed within the motor housing is circulat internal fan, so transferring the heat generatec windings to the housing. This heat is then transl the atmosphere by a second fan circulating free a the outside of the motor housing. For detailed information on electric motor s reference should be made to Australian Stan1 Fig. 12.1 The component parts of a 415 V, 3.7 kW, four-pole 1359 on the requirements for rotating e three-phase induction motor. This particular machines. It is an extensive standard with many motor is of the totally enclosed type and is and often calls up other standards that may be intended for direct coupling to its load as shown by the flanged construction of the endshield at to particular sections. Electrical rotating macl the upper left. POPE ELECTRIC MOTORS now classified by two letters followed by four n 222 THREE-PHASE INDUCTION MOTORS 223

I II \I II II \I II Fig. 12.2 Squirrel-cage rotor for an induction motor 111111 POPE ELECTRIC MOTORS 111111 111111 This classification number is different for such categories as cooling, mounting and protection. Fig. 12.3 Wound rotor for an induction motor 12.2.3 Rotor Squirrel-cage rotor The rotor of a three-phase motor consists of a shaft with poles. Usually the rotor winding has three phases, bearings, laminated iron core, and rotor conductors. The connected internally in star, and terminating at three slip most common type of construction is that with rotor bars rings. A typical wound rotor is shown in Figure 12.3. in the lamination slots rather than a winding. The rotor The slip rings are connected by means of brushes to a bars, short-circuited at each end by a solid ring, are often star-connected variable resistance, as in Figure 12.4. This made of copper strip welded to copper rings, but for small rotor rheostat provides the means of increasing the to medium size motors they may be cast in one piece out of resistance of the rotor circuit during starting, thereby aluminium. Usually included in the rotor casting is a series producing a high starting torque at a low starting current. of vanes for creating air movement. Figure I 2.2 shows As the speed increases, the external resistance is gradually these vanes standing out from each shorting ring. The reduced, lowering the rotor circuit resistance as the rotor photograph also shows skewed conductors in the rotor. reactance decreases. The main purpose for slanting the conductors in the rotor Under operating conditions, the variations in rotor is to ensure a smooth steady acceleration during starting. circuit resistance provide a means of cbntrolling the speed Varying the physical design features of the bars affects the of the motor-an increase in resistance produces a motor performance. Embedding them deeper into the reduction in speed. This also produces a loss in efficiency rotor, for example, increases their inductance and gives a due to the 12 R losses in the rheostat. lower starting current but at the same time creates a lower The wound-rotor motor is more expensive than the pull-out torque. squirrel-cage motor due to the cost of manufacture of the This type of rotor is then restricted to loads requiring wound rotor. It also has a higher starting torque and low-starting torques such as centrifugal pumps. The rotor lower starting current, but poorer running characteristics windings, if assembled without the laminations, resemble than the squirrel-cage motor. a metal cage giving rise to the often-used name of "squirrel-cage" rotors although the standards refer to them simply as "cage" rotors. 12.3 Operating principles 12.3.1 Rotating magnetic fields Wound rotor For its operation a three-phase induction motor is The wound rotor is fitted with insulated windings, similar dependent on a rotating magnetic field being established to the stator winding and having the same number of by the a.c. windings. The three separate windings are

Three-phase supply

Stator Rotor Slip rings Rheostat

Fig. 12.4 Circuit for a wound-rotor induction motor 224 ELECTRICAL PRINCIPLES FOR THE ELECTRICAL TF

A

+ A

..- ~ / / s, / .. 7 •·· .. I ··. I I .. I I \ I ·. / \,.. ,;:: / •/ .... / ccr ... · ......

(a) One phase (b) Three phases

Fig. 12.5 Polarities and connections in a two-pole, three-phase motor

installed in the stator at 120° E intervals to each other and for example, alternates in direction in the diagran provide a fixed number of poles for each phase. This is not rotate in any way. lt simply varies in stren shown diagrammatically in Figure l 2.5(a) for one phase direction in the vertical plane. Similarly a pulsatir of a two-pole machine. Figure 12.5(b) shows the three also established by the other two phases giving a phases in relationship to each other giving a total of six three magnetic fluxes which combine into one r poles. Phase A is drawn as a solid line, phase Bas a dotted flux. This flux rotates at synchronous speed. At r line and phase Casa dashed line. Note that this sequence is carried through for the explanation and applies to the current waveforms, the magnetic fields and the phasors. '• le Assun1ption ---. In the following explanation for the production of a \ / \ rotating field one assumption has been made as a ·; reference, that winding ends A, B, C when connected to a I ·.. \ \ I positive source of voltage makes the adjacent iron core a I I north magnetic pole. From this it will follow that the I opposite poles become south magnetic poles. These 0° : 120~ 240° \ I details are also shown in Figure 12.5(a). If the current \ . I flow is reversed then the magnetic poles are also reversed. \ With the three windings connected in star by joining ... \ I / ends Ai, B1, C, together, and the ends A, B and C ..... connected to a three-phase supply, the phase currents IA, ' !Band !care 120 °E out of phase with each other. These are shown in Figure 12.6. Because each current is 2 3 4 5 6 7 alternating, each pair of poles sets up a magnetic flux that Fig. 12.6 Waveform diagram showing three-phase< continually changes from one polarity to the other. Note at 120° E to each other (for reference num that although the flux set up by A phase in Figure l 2.5(b ), text) THREE-PHASE INDUCTION MOTORS 225

c ···· ...... / --,-,--••• -~- R ,/ ··· ...... B B

Flg.12.7 The resultant flux produced by currents flowing at position 1 in Figure 12.6 position I in Figure 12.6, the current IA is zero and no flux position I. If drawn to scale it can also be shown that the is produced by the winding A-A1. Current Is is negative length of the resultant has remained constant, indicating and so will produce a south pole at Band a north pole at that the field strength has remained constant. B1. Current leis positive and so will produce a north pole At position 3 (Fig. 12.6), IA is positive, producing a at C and a south pole at C1. Because currents /oand le are north pole at A and a south pole at A,, 18 is zero, and equal the two magnetic fields are equal in strength. The le is negative, producing a south pole at C and a north direction of these fields are shown in Figure 12. 7. In the pole at C1. These fields are drawn out in Figure 12.9 accompanying phasor diagram the addition of these two together with their phasors. The resultant field has rotated fields is shown giving a resultant instantaneous field 4>n. a further 60°E in a clockwise direction. (There is a 60°E At position 2 in Figure 12.6, IA is positive, Io is still difference between all the numbered positions in Fig. negative while leis zero. This produces a north pole at A, 12.6.) For each of the numbered positions the resultant a south pole at B, and nothing at C. These are shown in field rotates a further 60°E in a clockwise direction. For Figure 12.8 together with the phasor diagram showing the one complete cycle of current (360°E) the resultant addition of the phasors to give the resultant instantaneous magnetic field rotates 360°E. magnetic field. Since all coils have an equal number of turns, the relative strengths of the magnetic fields can be gauged by measuring the vertical heights of the current 12.3.2 Rate of rotation waveforms at the positions indicated by the reference BycomparingFigures 12.6, 12.7, 12.8and 12.9itcanbe number. In this instance the direction of the resultant seen that for the time intervals of 60° E between the magnetic field has shifted 60° E clockwise from that in positions l, 2 and 3 the resultant field rotates an equal

Fig. 12.8 The resultant flux produced at position 2 in Fig. 12.9 The resultant flux produced at position 3 in Figure 12.6 Figure 12.6 226 ELECTRICAL PRINCIPLES FOR THE ELECTRICAL Tl amount around the stator. For a complete cycle of a.c., a .------{_) R two-pole field rotates one complete revolution around the stator. The synchronous speed of the magnetic field in revolutions per minute can be determined from the frequency of the supply. ~c, Example 12.1 A two-pole machine is connected to a 50 Hz supply. Find c the speed at which the magnetic field rotates around the stator...... ow 50 Hz 50 cycles per second = I speed of rotation = 50 revolutions per second I = 50 x 60 revolutions per minute L _ ------0 B = 3000 r/min. (a) RWB sequence

With a four-pole machine, 360°E represents one-half of a full revolution of the stator field, and the speed of rotation of the field is consequently halved. Similarly, the speed of field rotation for a six-pole machine is reduced to one-third that of a two-pole machine. In each case the speed is usually expressed in revolutions per minute, whereas the frequency is in hertz (cycles per second). The speed in revolutions per minute can be found from the following formula: c B, .------~ ... ····· ' 120{ ' \Ow ns1•11. =-"-p \ ... :\ where nsyn = number of revolutions per minute ...... / bs .f = frequency in Hz (b) RBW sequence p = number of poles The speed n of the rotating magnetic field is called the Fig. 12.10 Phase sequence and field rotatil synchronous speed of the motor. The synchronous speeds of common sizes of motors at a frequency of 50 Hz are 12.4 Induction and its effects given in Table 12.1. The formula above is identical to that When the stator windings of a three-phase i shown in Chapter 6. motor are energised from a three-phase s1 magnetic field is produced, rotating at sync Table 12.1 Speed of the rotating field in an induction speed. This rotating magnetic field crosses the ai1 motor for various number of poles cuts the rotor conductors, inducing a voltage Poles 2 4 6 8 1 O 12 (magnetic field, conductors and relative motior Synchronous the rotor circuit is complete (through end rings it speed (r/min) 3000 1500 1000 750 600 500 of the squirrel-cage rotor, or external resistance ii of the wound rotor), the induced voltages ca currents to flow in the rotor conductors. 12.3.3 Direction of rotation and reversal The direction of rotation of a rotating field depends on the 12.4.1 Torque phase sequence of the three currents flowing through the Figure 12.11 (a) represents a part of the stator an windings. In Figure 12.1 O(a) the three supply lines R. W of an induction motor with the stator flux rot2 and Bare connected to terminals A, Band C of the motor. clockwise direction as indicated. When these line The resultant magnetic field rotates in a clockwise cut the rotor conductors from left to right, the direction. movement between the stator flux and the re In Figure 12. 1O(b) the supply lines to phases B and C ductor is from right to left. By applying Flemin have been changed over and. using the procedure from the hand rule (sect. 6.1. l) the direction of inducec previous section. it can be shown that the rotation of the flow in the conductor is towards the reader. D magnetic field is reversed. That is. the direction of rotation comparatively high rotor currents flowing, a Jar of the field. can be controlled by interchanging any two established around the conductor as sho\vn i supply lines to the motor. In section 12.4.1 on torque it 12.1 l(b). The stator and rotor fluxes react withe; is shown that the rotation of a three-phase induction as shown in Figure 12.11 (c) to form a resultant f 111otor is in the same direction as that of the rotating field. resultant field tends to straighten itself out, a1 THREE-PHASE INDUCTION- MOTORS 227 N LJ N Rotor conductor -Thrust ~Rotor flux

' . (a) Stator flux {b) Rotor flux (c) Resultant flux Fig. 12.11 Production of torque in an induction motor process causes a force to be exerted on the rotor The formula for determining percentage slip is: conductor trying to force it to the right and out of the stator magnetic field. A similar force is exerted on all the rotor conductors as the field rotates, and if sufficient force s% = nsyn - n x 100 is created the rotor will commence rotating in the same nsyn direction as the rotating magnetic field. Provided it is free to rotate the rotor will accelerate until it approaches where s% = percentage slip synchronous speed. nsyn = synchronous speed This rotating force, called the torque of the motor, is n = rotor speed the result of the interaction of the two fluxes. The stator flux remains fairly constant, but the rotor flux varies with At standstill (i.e. when starting) the slip is !00%, the rotor current, which is determined by such factors as whereas if the motor could run at synchronous speed, the the impedance, the induced voltage and the relative speed slip would be zero. of the rotor conductors.

12.4.2 Slip 12.4.3 Rotor frequency To produce torque, there must be a rotor flux caused by When the rotor of a two-pole motor is at standstill and the current flowing through the rotor conductors. If the rotor stator is connected to a 50 Hz supply, each rotor could run at synchronous speed, there would be no conductor is cut by a north pole and a south pole at a rate relative motion between the stator flux and rotor of 50 times per second. At standstill, the frequency of the conductors. Consequently, there would be no induced rotor voltage (rotor frequency) is the same as the voltage, no rotor current, no rotor flux, no torque frequency of the supply (stator frequency). developed, and so the rotor would slow down. An induction motor therefore cannot run at synchronous As the rotor speeds up to half the synchronous speed speed. (!500 r/min), the rotor conductors are cut by only one­ With the rotor runningjust below synchronous speed, half as many north and south poles per second as at relative motion exists and sufficient torque is developed to standstill, and so the rotor frequency is one-half the keep the rotor turning. The difference between the supply frequency (i.e. 25 Hz). If the rotor revolved at synchronous speed of the rotating field and the actual synchronous speed, the rotor frequency would be zero. speed of the rotor is calfed the slip speed. It is commonly The rotor frequency depends upon the differences in the expressed as a percentage of the synchronous speed. speeds of the stator flux and the rotor (i.e. the slip of the motor), as shown in Figure 12.12. Example 12.2 Determine the slip of a four~pole induction motor running The rotor frequency can be calculated using the at 1440 r/min when connected to a 50 Hz supply. following formula: 120{ 120 x 50 = p = = 1500r/min n,,. 4 ~ slip speed = 1500 - 1440 = 60 r/min ~ 1500 44 percentage slip ~~ 0 x 100 where fr = rotor frequency in Hz 1 s = slip percentage = 4% f = supply frequency in Hz 228 ELECTRICAL PRINCIPLES FOR THE ELECTRICAL TF

Supply frequency the difference between its speed and the rotating m ...... field becomes less and the generated voltage cam rotor circulating currents also becomes less. This reduces the stator current. The typical relationship between the stator Rotor and the rotor speed is shown in Figure 12.13( frequency initial circulating current in the rotor is affected frequency of the supply, the resistance of the rot and the inductance of the rotor circuit-that is, the limiting factor is the impedance of the rotor circu /Rotor stationary the usual type of power transformer, the frequ the supply is the line frequency, but in this c 0 Slip 100°/o frequency commences at line frequency and Fig. 12.12 Relation between rotor frequency and slip decreases as the motor speed increases. As a consi the torque created can change as the speed chan. Example 12.3 Figure l 2. l 3(b) for the typical relationships I Determine the rotor frequency of a two-pole, 50 Hz speed and torque. For small values of slip the t induction motor if the rotor speed is 2850 r/min. assumed to be proportional to the slip. As the mo increases the torque increases and the speed de s = nsyn - n x 100 nSJ'll until the torque reaches a maximum value ca 3000 - 2850 breakdown torque. If the motor is loaded bey' 3000 x 100 = 5% point, the torque and the speed both decrease motor quickly comes to a standstill. An overa s.[ f,, =TOO for starting torque is in the region of 1.5 times t torque, while the breakdown torque is usually abc 5 x 50 the rated torque. Australian Standard 1359.41 2.5 Hz = loO = minimum requirements for these torque val· As the rotor frequency varies, so does the rotor provides a table for a range of motor sizes. inductive reactance, and this affects the starting and The resistance of the rotor conductors running characteristics of the motor. constant at power line frequencies for all 1 purposes, while the inductive reactance decreas( rotor speed increases. As a guide, torque re 12.5 Operating characteristics maximum when the rotor resistance in ohms is the rotor reactance in ohms. Since the resistance 12.5.1 Squirrel-~age motors then the breakdown torque can only be al When power is first applied to a stationary motor, the relationship to the motor speed by altering the inc stator windings act as transformer primary windings with of the rotor. In turn this affects the starting the resultant magnetic field rotating at synchronous Australian Standard 1359.41 allows for only t• speed. The rotor then behaves as a shorted secondary types of rotor-normal and high torque-and a winding causing a high circulating current in the rotor types are necessarily by prior arrangements bars and a high starting current in the stator windings. As 1nanufacturer. For the high-torque motor the the rotor accelerates in the direction of the rotating field, breakdown torque remains around twice rate<

Breakdown torque

Current (0/11) Torque(%) Locked Rated )._1or torque speed

Rated torque 100 Rated current 100

0 n fl syn 0 n n, (a) Current/speed curve {b) Torque/speed curve

Fig. 12.13 Operating characteristics tor a squirrel-cage induction motor THREE-PHASE INDUCTION MOTORS 229

300 300 r---- 200 Torque 200 % Torque Rated torque % Rated torque 100 100 ------

Oc______~-l- Speed n~yn Speed Fig.12.17 High resistance rotor bars Fig. 12.14 Low starting torque rotor bars bars of greater cross-section where part is imbedded deeper into the rotor magnetic circuit. Starting torque is still about 150% of running torque but the starting current is reduced to about five times the running current. It is suitable for use with equipment of low starting inertia such as fans, blowers and some types of machinery. Figure 12.16 gives one example of a rotor with two sets of rotor bars. The inner set is shown with half as many 300 Rated bars as the outer set and includes an optional air gap. speed Depending on performance requirements there may 200 Torque be different shaped bars, no air gap or a full set of bars in % Rated torque 100 the cage. Starting torque is high-here it is 225% of rated torque-and starting current is about five times rated running current. Applications are air compressors, o~------e~ Speed nsyn crushers, refrigerator compressor motors or reciprocating force pumps. A typical example of high resistance rotor Fig.12.15 Standard rotor bars bars with low starting current requirements is shown in Figure 12. l 7. With this construction the starting torque can be increased to about 275% with fairly low starting currents. It is at the expense of a lower rated speed (i.e. increased slip). Typical uses are flywheel mounted machinery such as presses and punches. It is excellent with hoists where the maximum load occurs at the start of the lift. The details above apply particularly to copper rotor bars. If aluminium is used for the rotor bars then the cross-sectional area of the bars must be increased to allow 300 for the metal's higher resistivity. The shape may also be ------.,Rated _____ speed changed to incorporate desired starting and running 200 Torque characteristics. Figure 12. l 8 shows a "tear-drop" shaped % Rated torque cast aluminium bar. In practice the shape may also be 100 ------inverted to alter the characteristics to suit a particular purpose. o~------'+- Speed nsyn

Flg.12.16 Double cage rotor tor high starting torque while the starting torque is increased to approximately 2.5 times rated torque. Figure 12.14 shows typically shaped rotor bars where the starting current is about 6-7 times rated current and the motor has a starting torque of approximately 150% of rated running torque. Its use is restricted to very low starting torque requirements. Figure 12.15 shows rotor Fig. 12.18 Cast aluminium rotor bars 230 ELECTRICAL PRINCIPLES FOR THE ELECTRICAL T 12.5.2 Wound-rotor motors The normal starting procedure is to start th The introduction of resistance into the rotor circuit of an with all the resistance in the rotor circuit. As th induction motor produces three effects: speeds up the resistance is reduced and tht increases in speed, but maintains a high torque. I. rotor current is reduced, resulting in less stator current; the starting procedure, the torque-speed curve is c 2. starting torque is increased, because rotor and stator by the thicker curve d. magnetic fields are more in phase with each other; 3. slip speed is increased. 12.5.3 Operating parameters An adjustable resistor is used external to the rotor, By comparing Figures 12.13 and 12.19, it can be' which is wound with comparatively low resistance full-load torque occurs at a greater slip in a wow windings. The value of the external resistance can be motor than a squirrel-cage motor. This is due to 1 adjusted as required and as the motor accelerates, the resistance of the windings in the wound rotor. value is gradually reduced until all the resistance is out of On no load, the stator current ofany inductic the rotor circuit and the motor behaves as an ordinary is largely a magnetising current, with a smal induction motor. component required to supply the no-load The torque-speed characteristic of a typical three­ Accordingly, the power factor of an induction n stage wound-rotor motor is shown in Figure 12.19. When no load is very low. The no-load current is relatb all the resistance is in the rotor circuit, the starting current when compared with a transformer because of is low and the starting torque is high as shown by curve a. reluctance of the magnetic circuit, due to the If this resistance is left in, the full-load torque would occur between the stator and rotor. at approximately 25% slip, resulting in extremely poor The stator flux remains fairly constant from n speed regulation. full load, and so the magnetising current is al If one stage of the resistance in the rotor circuit is constant. In Figure l 2.20(a), the no-load stator c shorted out, the operating characteristics are modified as lags the supply voltage by cf> degrees. shown by curve b. As a load is applied to the motor, a load cun If all the Cxternal resistance in the rotor circuit is required to accommodate that load. This load ct shorted out the operating characteristic is shown by lags the supply voltage slightly, due to the effo curve c. stator and rotor reactance. The two current corr Io and /'1 combine to give the total stator current load. The phase angle decreases from cf> to q,, power factor of the induction motor increases as on the motor increases. Figure 12.21 gives representative characterist d for some parameters of a three-phase induction 1 shows the speed decreasing and the slip increasi: Rated speed load on the motor is increased. It also shows r' current increasing and the power factor improvi same time . ... ······ a Aated_!_o~~ ___ _ 100 12.6 Motor starting methods Induction motors are subject to the same limitations as d.c. motors. In Chapter 7 it was sh 0 n a d.c. motor needed a series resistor to limit f starting current and, as the motor accelera Flg.12.19 Operating characteristic for a wound-rotor motor resistance was gradually reduced until it was no

---- (a) No-load conditions (b) Loaded conditions Fig. 12.20 Current phasors for an induction motor THREE-PHASE INDUCTION MOTORS 231

100% L, 0-----j

L'D-----1

Lau-----1

Fig. 12.22 Basic power circuit for primary resistance motor st·arting

resistance, the greater is the voltage drop across each resistor and the less the voltage at the motor. Because of this lower voltage, the starting current is reduced. As the rotor accelerates, the resistance is reduced in steps until Power output full voltage is applied across the motor terminals. This method of starting greatly reduces the starting Fig. 12.21 Typical parameters for a three-phase induction motor showing how they alter as the load varies torque of the motor because it is proportional to the square of the applied voltage. circuit. Limiting the current drawn by the motor during starting is directly applicable to both a.c. and d.c. motors, i.e. and is achieved by reducing the voltage across the motor It must be appreciated that this expression shows the windings. The only exception to this is the wound-rotor motor. starting torque is reduced four times if the applied voltage is halved. Such a reduction in torque may prevent a motor While ad .c. motor can only use series resistance, there 1 from starting against even a small load. are alternative current-limiting methods for a.c. motors. Advantages of current limiting 12,6,3 Star-delta starting I. Less mechanical forces exerted on windings. Another way to reduce the starting current by reducing 2. Less mechanical shock forces on motor frame, shaft the voltage applied to each winding is the star-delta and transmission. method. The motor is started in the star connection and 3. Steadier acceleration with connected loads. when it has gained sufficient speed, it is quickly changed 4. Reduction of line voltage drop. over to the delta connection. 5. Less disturbance to the supply system. The starting torque is considerably reduced with this method, which is usually applicable to motors being Disadvantages of current limiting started on light loads, the main load being applied after I. Reduction in starting torque. the motor has reached full speed and is connected in delta. 2. Extra cost of starting equipment. The two ends of each phase winding of the motor are 3. Increased maintenance requirements. brought out to the stator terminals. During the starting 4. More con1plex equipn1ent. sequence in star, the voltage across each phase is 1/ V3 or 58% of the line voltage. As a result, the torque is reduced 12,6,1 Direct-on-line starting to ( 1/ VJ) 2 or 1/3 of its normal running value. Cage motors may be started with the full supply voltage The line and phase currents in star are equal, but when connected across the stator windings. This method is the windings are connected in delta this condition no usually referred to as direct-on-line or D.O.L. starting. longer applies. The phase voltage is increased VJ times or The large starting current can cause excessive voltage 173% over the star connections, consequently the phase drop in the supply lines and disturbances to the supply current is increased by the same ratio. In addition the line voltage. For this reason, the supply authorities usually current is now equal to V3. Iphase, or three times the line limit the starting current of motors. The D.0.L. method current value for the star connection. These values are of starting is usually restricted to smaller-sized motors. illustrated in Figure 12.23 for a winding impedance of 24 ohms. 12,6.2 Primary resistance starting An effective method for reducing the starting current of an 12.6.4 Autotransformer starting induction motor is to add resistance in series with the Autotransformer starters are the most popular of any supply lines (see Fig. 12.22). The higher the value of reduced voltage type. The voltage applied across the 232 ELECTRICAL PRINCIPLES FOR THE ELECTRICAL 1

Torque and input power 10 A 113 of "run" condition 30 A

10 Ai

Start (star-connected) Run (delta-c

Fig. 12.23 Comparison of star and delta starting stator windings can be reduced to a percentage of the Example 12.4 supply voltage by using a star-connected autotransformer. A 415 V, three-phase induction motor draws 160 During starting, the primary of a step-down auto­ connected D.O.L. If an autotransformer starter, transformer is connected to the supply, and the secondary motor connected to the 70% tapping, is used to is connected to the stator windings of the induction motor, determine: motor, as shown in Figure 12.24. (a) the voltage applied to the motor during start By providing a range of tappings in the transformer (b) the starting current taken by the motor; windings (e.g. 60%, 70%and 80%), it is possible to have a (c) the starting current drawn from the supply. choice of voltages (and currents) for starting purposes. When the motor is up to speed, the stator windings are ,, connected across the full supply voltage, and the autotransformer is open-circuited. The use of a transformer makes it possible to reduce ,, the line input current at a greater rate than that at which V1 = the torque is reduced. Transformers are discussed in 415 v greater detail in Chapter 15, but briefly: Vi = 70°/o of V1 input voltage x input current = output voltage x output current Fig. 12.25 Circuit diagram for example 12.· i.e. v, x r, = v, x r, (neglecting all losses) (a) For 70% tapping: During starting, a reduced voltage ( V,) is applied to the motor, so reducing the starting current(!,). Because of motor voltage = 70% of input voltage transformer action, however, the input current (/1) is =70%of415V reduced still further. It can be illustrated by the following = 0.7 x 415 example. = 290.5 v

0

Fig. 12.24 Starting connections with a three-phase, star-connected autotransformer THREE-PHASE INDUCTION MOTORS 233 (b) For 70% tapping: motor current = 70% of D.O.L. starting current = 70%of 160A = 0.7 x 160 = 112A

(c) V,J, V2l2 :. J, V,h = 290.5 x 112 v, 415 = 78.4 A Fig. 12.26 Starting connections for an induction motor using an open-delta autotransformer Note that the motor voltage has been reduced by 70% 2 to 290 V. Because T o:: V , the torque will have been 12.6.5 Secondary resistance starling reduced to (0.7)2 =0.49 of the D.O.L. value. While the motor current has only been reduced to 70%, the input This method of starting can only be used with wound­ current has been reduced to 0.49 or 49% of the D.0.L. rotor motors. Full line voltage is applied to the stator value. windings and the starting current is limited by connecting external resistance across the rotor terminals, as shown in The significance of these figures can be seen when Figure 12.27. As ihe motor's speed increases, the compared to those of the same motor when the primary resistance is gradually removed from the circuit until at resistance starting method is used. full speed all the resistance is shorted out. Motor impedance between lines at standstill 12.6.6 Other types of motor starters v 415 z =I= 160 = 2.59 !1 (D.0.L.) Liquid Instead of using resistors as a form of current limiting, The line input current from example 12.4 is 78.4 A one trend is to use liquid containers with two electrodes (autotransformer). and a chemical electrolyte. Although this is often more For primary resistance starting with this same value of compact than the resistor type there needs to be some form of increased maintenance to check on the state of current the applied voltage would be reduced to: the electrolyte. The liquid is a simple replacement for V = !Z the resistor and is often a mixture of water and salt, with = 78.4 x 2.59 occasionally other chemicals. =203.lV Solid state For the same input current for both methods (78.4 A), Starters with this type of construction are initially the relative torque values would be: expensive, but generally incorporate some form of operator-adjustable starting current control. In addition, 415)' D.O.L. (ill x 100 = !00% many systems incorporate a variable frequency generator for speed control. Most models work on a principle of 2 5 converting the alternating current to direct current and autotransformer )' x 100 = 49% ( !~5 then generating voltages and frequencies to suit. See also sections 12.9.2 and 17.9.2. 203 · 1 )' 100 = 24"' primary resistance ( 415 x 70 12. 7 Typical pushbutton-operated That is, for the same input current the auto­ transformer starter enables the motor to develop twice as starter circuits much starting torque as the primary resistance method. 12.7.1 Circuit protection With three autotransformers used as in Figure I 2.24 it Electrical starting circuits are protected against faulty is usual, when changing over to full voltage, to open­ operation with fuses, circuit breakers and contactors. circuit the star-point and momentarily supply the motor through part of the transformer windings in series. These parts of the windings are then shorted out, effectively L, 0-----,' taking them out of the circuit. This method is called the Korndorfer method of starting. Autotransformers are L2 0----; more often used in the open-delta circuit where only two windings are used as in Figure 12.26. It is a cheaper Lo o------', method and while the circuit is unbalanced during the starting sequence, it is balanced as soon as the motor is in Fig. 12.27 Basic power circuit for secondary resistance the running connection. motor starting 234 ELECTRICAL PRINCIPLES FOR THE ELECTRICAL - Fuses L, A fuse is designed to become an open circuit once a certain L, value of current is exceeded. This does not mean that a La fuse will "blow" immediately its rating is reached. The fuse element can carry small overloads for a period of time, depending on the amount of the excess current and the rate at which the heat being generated can be dissipated. This characteristic makes a fuse suitable for motor protection circuits, where the fuse must be able to handle the starting currents. * Circuit breakers Circuit breakers, on the other hand, can be designed to operate with only small overloads and steps may have to be taken to slow the action down and enable motors to be K1.3 E started. Some circuit breakers operate on a magnetic K1.1 K1.2 attraction principle, others on a thermal element; most operate with both magnetic and thermal elements.

Contactors Oil- Both fuses and circuit breakers are designed to protect electrical circuits against excessive currents, but serve no useful purpose in the event of power failures. As a means of protection contactors are used. When a power failure or low-voltage situation occurs the contactor drops out, so switching the equipment off until power is restored. As an added protection against faulty starting sequences, Power circuit I Control ci. most motor starters are automatic once the initial pushbutton operation has been made. Fig. 12.28 Contactor circuit tor D.0.L. starting The following examples show pushbutton-operated circuits for each of the five means of starting three-phase induction motors. The circuits are shown with fuse protection and thermal overload current protection as possibly the most common protection methods en­ L, countered. Individual manufacturers have their own L, preferences for starter circuits and these may vary in detail from one firm to another and from one model to another. La

12.7.2 D.O.L. contactor starter circuit (Fig. 12.28) Circuit operation

1. Pressing the start button completes a circuit from L 3 through the normally closed stop button to coil KI, and the overload to L2 • K1.1 2. Main contactor coil Kl then closes and applies full line voltage directly to the motor via contactor contacts Kl.I, Kl.2 and Kl.3. 3. Contact Kl.4 bridges out the start button contacts so that, on the release of the start button, the contactor remains in the operational state-Le. the control circuit is latched in the "on" position. Pressing the stop button disables the latching circuit and allows the main contactor to revert to the "off" state.

12.7.3 Primary resistance contactor starter circuit (Fig. 12.29) Circuit operation Power circuit Control ci1 1. Pressing the start button completes a circuit from L3 through the normally closed stop button to coil KI, and the overload to L2. Fig. 12.29 Contactor circuit for primary resistance s THREE-PHASE INDUCTION MOTORS 235

2. The main contactor KI operates. Contact Kl .4 closes 2. When K2 operates, it causes the "ends" of the three and bridges out the start button contacts, so that on windings to be joined in "star" via contacts K2. l, K2.2 the release of the start button the Kl contactor circuit and K2.3. remains latched. 3. Simultaneously, coil K3 is open-circuited by K2.5. 3. Contacts Kl.I, Kl.2 and Kl.3 close, and a reduced This is the delta connecting coil and must be isolated line voltage is applied to the motor through the when the star connection is in operation. Similarly, resistors in series with each line to the motor. The when the delta connection is in operation the star starting current is limited by the resistors to a value connection must be isolated, a method called below that of D.O.L. starting. "electrical interlocking". As a precaution, the star and 4. Delayed action contact K 1.5 operates after a predeter­ delta connecting contactors are often mechanically mined delay and completes the circuit for coil K2. Its interlocked in addition to the electrical interlocking operation causes contacts K2. l, K2.2 and K2.3 to provided by contacts K2.5 and K3.4. close and allow full line voltage to be applied to the 4. When K2.4 closes, a voltage is applied to the timer motor. K4 and to coil Kl. This allows Kl.4 to close and 5. Pressing the stop button de-energises all coils and bridge the start button. allows the starter to revert to the "off' state. 5. Contacts Kl.I, Kl.2 and Kl.3 close and apply a voltage to the "starts" of the motor windings. 6. The voltage applied across the windings is only a 12.7.4 Star-delta contactor starter circuit proportion of full line voltage (0.58), and starting current is reduced accordingly. (Fig. 12.30) 7. When the time delay period has elapsed, contact K4. l Circuit operation opens and forces contactor K2 to disconnect the star I. Pressing the start button completes a circuit from L3 connection. The dropping-out action of K2 completes through the normally closed stop button and two the circuit of coil K3 through K2.5, and it is then normally closed contactor contacts (K4.l and K3.4) activated. This open-circuits the interlocking contact to coil K2, and the overload contact t to L2. K3.4 and also switches off the timer K4 via K3.5.

L, L,

E

E

K3.1 K3.2 K3.3 K4.1 1<1.4 I I ~ K3.4 K2.5 K3.5

K2/5 K3/5

Control circuit

Fig. 12.30 Con/actor circuit for star-delta starting 236 ELECTRICAL PRINCIPLES FOR THE ELECTRICAL - 8. Contacts K3.l, K3.2 and K3.3 close and complete the voltage is available to the control circuit whe delta connection, allowing full line voltage to be button is released. applied to the motor. 5. The delayed opening contact Kl.5 opens afi 9. Pressing the stop button de-energises all coils and determined time lapse and forces K2 to op allows the starter to revert to the "off' state. the delta connection. Contact K2.4 then clo and coil K3 is activated. 6. Contacts K3.l and K3.2 close, and full lin 12.7.5 Autotransformer contactor starter circuit is applied to the motor through the Kl cc (Fig. 12.31) series with two lines. The electrical interloc Circuit operation K3.3 opens and isolates coil K2. 7. Pressing the stop button de-energises all 1. Pressing the start button completes a circuit from L 3 allows the starter to revert to the "off'' state. through the normally closed stop button, a normally closed delay contact Kl.5, electrical interlock K3.3, 12.7.6 Secondary resistance contactor start coil K2, and the normally closed thermal overload (Fig. 12.32) contact t to L2. 2. When K2 is activated, it closes the contacts K2. l and Circuit operation K2.2 connecting the ends of the autotransformers to l. Pressing the start button completes a circui1 line L2 in an open delta configuration. through the normally closed stop button, coi 3. The operation of K2 simultaneously closes contact the thermal overload contact t to L2• Coil ~ K2.3 and opens contact K2.4-the electrical interlock is in parallel with coil Kl, is activated at to prevent K3 operating while K2 is active. time as Kl but only operates after a pred< 4. K2.3 supplies power to coil Kl, which is also activated. time delay. Contacts Kl.I, Kl.2, Kl.3 and Kl.4 close. Full line 2. Contact K 1.4 bridges out the start button cc voltage is connected to the autotransformers and a that on the release of the start button the , reduced line voltage is supplied to the motor via the remains in the operational state-Le. the cont1 transformer tapping. Contact Kl.4 ensures that a is latched in the "on" position.

L, L, La I I I E K1 .1 K1.2 K1.3 I IE K1.4 I K3.1 K32 K2.3 1 I I I I ~ I Power circuit I Control circuit Fig. 12.31 Contactor circuit for autotransformer starting THREE-PHASE INDUCTION MOTORS 237 L, L, L,

K1.4

E

K2.1 K3.1

R R R K1/4 K4/2

R R R

Power circuit Control circuit

Fig. 12.32 Contactor circuit for secondary resistance starting

3. Contacts Kl.I, Kl.2 and Kl.3 close and apply full 12.7.7 Part winding starters line voltage to the stator terminals of the motor. The Another method of motor starting gradually gaining a rotor has two resistors in series with each winding and, measure of acceptance is the part winding motor starting as the ends are connected in star, current flows in the method. As with other methods the primary intention is rotor windings and the motor is able to generate torque to reduce starting current and/or torque of a motor. With and commence turning. larger motors the initial starting torque can transmit high 4. After the delay time, K2 operates and closes contact and damaging shock values to transmission components, K2.l. This causes K4 to be activated along with K3, and some type of reduced torque starting becomes a second time delay relay. essential. 5. Contacts K4.l and K4.2 then close and reduce the The phase windings of the stator are divided into amount of resistance connected across the slip rings. parallel sections each of the requisite number of poles This action enables the motor to attain a higher speed. and each capable of withstanding full line voltage. Parts 6. After a further time delay, coil K3 operates and closes of the stator winding are energised and as the motor gains contact K3.l. Coil K5 is then activated and closes speed more sections of the stator winding are energised. contacts K5 .I and K5 .2. This action removes the Normal control and power components are used to remainder of the resistance in the rotor circuit and provide the necessary switching. As much of the motor's the motor is in its normal running mode. windings remain connected to the line in a closed 7. Pressing the stop button de-energises all coils and transition sequence, current surges are kept to a allows the starter to revert to the "off' state. minimum. 238 ELECTRICAL PRINCIPLES FOR THE ELECTRICAL l

Table 12.2 Relative characteristics of various starting methods Starting Stator Starting Starting No. of Current surge Types of Example loads Genera method voltage at current torque starting during loads comme 0 start ( /o/FL) {O/o TFL} steps transition suited stages Direct-on- line voltage 700°/o 150o/o n.a. light ine~ia centrifugal starting line loads pumps; lathes greate1 load to Primary reduced 300o/o 40o/o 2+ no almost no fans poor sta resistance load torque Star-delta reduced 200°/o 33o/o 2 yes light loads motor- starting generator 113 full I units torque Autotrans- reduced 300o/o 80o/o 2 yes substantial hydraulic starting former proportion pumps; slightly of full load conveyors than fu torque Secondary line voltage 1 OOo/o 1 OOo/o 2+ no high inertia shock loads; rotor ref resistance loads presses; adjuste shears start (T

Note: The figures quoted in this table must be considered as a general ~uide only. Many variables can be encountered-types an, stators and rotors, applications, loads and starters being only some oft e factors.

In Figure 12.33 each phase winding has been divided L, into two sections, each section sequentially being connected to the line voltage. Starting torque is down to approximately 45% and starting current is about 65% of normal D.0.L. starting. Because of current imbalance during the starting sequence, the motors tend to be noisy at this time, which restricts application of the methoc! to cases in which a motor requires occasional starting before running for long periods. (a) Start (stage 1)

12.8 Motor output power L, L, In an induction motor, the magnetic fields of the stator and rotor interact to cause a force to be developed on the rotor conductors. This force, acting at a distance from the shaft equal to the radius of the mean circle of the rotor conductors, develops a torque or turning force at the shaft of the motor. Torque is measured in newton-metres (Nm) and is calculated by the product of force (newtons) and (b) Transition (stage 2) radius (metres).

T= F.r L, L, L,

Example 12.5 A motor exerts a force of 360 newtons at the rim of a pulley with a diameter of 0.5 metre. Calculate the torque developed by the motor.

T= F.r (c) Run (stage 3) = 360 x .Q2 = 90 Nm 2 = Fig. 12.33 Part winding starting THREE-PHASE INDUCTION MOTORS 239 Example 12.6 12.9 Speed control of induction motors If the n1otor in example 12.5 was fitted with a 0.3 m diameter In section 12.3.2 the synchronous speed of an induction pulley, calculate the force exerted at the rim of the pulley. motor was shown to be: 120( F = .I_ nsyn = r p 90 That is, the speed depends on both the frequency and the - 0.15 nu1nber of magnetic poles in the machine. = 600N The synchronous speed is that of the rotating magnetic field, while the actual speed must always be less When considering the mechanical output of the than this by the amount of slip necessary to allow the induction motor, it is necessary to determine the power motor to develop the required amount of torque. To produced. change the speed of an induction motor by an appreciable amount, other than by loading it to alter the slip speed, power = rate of doing work = work done per second (watts) either the frequency of the supply or the number of poles in the windings must be changed. Under normal circum­ work done = force exerted x distance moved stances the speed of the induction motor must then be distance = 2rr.r for 1 revolution (where r = radius in metres) considered as fixed. If any application requires that the speed be adjustable, then special and expensive n = 2 rrr X for 1 second equipment must be used. (Refer to sect. 17.9.) 60 (wheren = r/min) 12.9.1 Wound-rotor motors power = force x distance The degree of slip in a wound-rotor motor may be 2rrrn 2rrn changed comparatively easily by varying the amount of =FX(;0=6QXFr external resistance in the rotor circuit. This provides dubious results in that the speed changes every time the But torque ( 7) = Fr load changes, efficiency can be as low as 40% because of the /ZR losses in the resistors and the available torque 2rrnT i.e. P =--watts can be appreciably reduced. In addition, heavy-duty 60 resistors have to be provided owing to the fact that the normal resistors are rated for starting purposes only (i.e. where n is in r/min. a short duty cycle). Combinations of wound-rotor motors connected in various ways have attempted to over­ Example 12.7 come these disadvantages-called cascading and A 415 V, three-phase, 50 Hz, four-pole induction motor concatenation-but the methods tend to be cumbersome has a full-load speed of 1440 r/min. Calculate the power and expensive. Special motors have been developed with produced by the motor if it develops a torque of 100 Nm. better results, but the costs are still so high that the direct current machine becomes competitive even after allowing p = 2rr.n. T 60 for the provision of a d.c. supply. 2 x .,,. x 1440 x 100 12.9.2 Squirrel-cage motors 60 This type of motor under normal operating circumstances 15 079 watts is considered as a fixed speed machine with only very 15.1 kW small variations in speed from no load to full load. Reducing the supply voltage has negligible effect on the Example 12.8 speed but reduces the amount of available torque and If the above motor draws 31 A from the supply, and the eventually the machine stalls. power factor is 0.86, determine the efficiency of the Special connection arrangements of the windings motor. 1hc1nsclves allow n1otors to be connected, for example, as either two- or four-pole motors. This only provides a step change in speed and allows no other control. If other P;. = VJVIA. speeds are required then possibly a second winding has to _ output be used. For example, one winding could give both two T/ - input and four poles while the second winding gives six poles. It = 15 079 x IOO is however still only speed changes in three steps. While VJVIA. satisfactory for some uses such as lathe work, it is not 15 079 x 100 suitable for any project such as rolling mills where = ,jj x 415 x 31 x 0.86 incremental changes may be required. The method of = 78.7% speed change in the ratio of 2: I for single windings was developed many years ago. More recently the advent of 240 ELECTRICAL PRINCIPLES FOR THE ELECTRICAL T

control of the operator, and can be done man1 a control knob on the unit. The speed range ol is from 1% to 200% of normal running speeds and efficiency are high throughout the complete The unit's operating principle is to convert ti phase alternating current supply to direct current to generate a three-phase supply at any desired f1 within the range of the equipment. The units 1 reversing facilities, and provision can be made fo control stations. The units are available in a sizes. The photograph of a typical unit is shown i 12.35.

12.10 Motor braking The main power equipment in most electrical sys rotating electrical machine. There are many app where it is necessary to bring the machine tc quickly, to keep a load from moving or to enable: reversal of direction. In general there are two braking-mechanical and electrical.

12.10.1 Mechanical braking Mechanical brakes are mostly solenoid operated tension holds the brakeshoes or band against ad it operates as a stopping device as well as a parkir This is necessary to prevent undesired movemen could occur with a crane holding a suspended loa power is applied to the driving motor, the soleno activated, so releasing the brake and allowing tr . . full control. Light duty brakes may adopt the di . IJ configuration but these have limited application:

12.10.2 Electrical braking Fig. 12.34 A two-stage secondary resistance starter. The There are two types of electrical braking and e; direct-on-line contactor is at the top, while the also be used in conjunction with mechanical bra lower contactor short-circuits the rotor resistors when operated by the timer. Dynamic braking In this method the motor converts its energy o pole amplitude modulation (PAM) represented an into electrical energy. For a three-phase inductic advance on this concept in that it gave closer ratios in the order of 6:4 and 8:6 and other similar ratios. Note however it is still step changes and not incremental. The PAM motor is made in all sizes and its "secret" lies in the internal connections of the pole groups. Six leads are brought out in a similar fashion to two-speed motor windings. The windings are halved and each half is connected in star across the supply lines, giving double or parallel star for one speed, while the windings are all connected together in series delta for the other. Another recent development for speed control of squirrel-cage motors is the electronic control unit. Provision is usually made in the unit for adjustable rates of current increase to control starting currents, and the voltage supplied by the unit to the motor is stabilised. The motor is accelerated from standstill at a predetermined current rate. Fig. 12.35 An incremental speed control unit for i Once running under load, the motor speed can be motors. It uses an electronically gener varied by changing the frequency. This is usually under variable frequency. ASI THREE-PHASE INDUCTION MOTORS 241 the most common method is to disconnect the a.c. supply Satisfactory operation of three-phase motors on a and reconnect the windings to a source of d.c., as shown three-phase supply depends on several factors: in Figure 12.36. An inspection of the circuit will show I. three equal voltages at the correct phase displace1nent. that the main and brake contactors are electrically Under normal operating conditions the phase interlocked with a time delay to switch the d.c. off after displacement is a function of the generating equipment a short period. Because the rotor conductors are still and stays relatively fixed, but the line voltages can rotating and the d.c. field is stationary, there is a vary depending on the individual loads connected at generated voltage and a circulating current in the rotor that tin1e. For balanced loads, such as three-phase bars. This creates a torque in the opposite direction to motors, unbalanced phase voltages lead to unbalanced the rotation of the n1otor, resulting in a rapid slowing currents flowing in the motor windings. As a down. The rate of slowing down is fairly constant, unlike consequence, circulating currents are set up, heating the d.c. n1otor with dynan1ic braking, where the braking is increased and uneven, and torque is reduced. effect din1inishes markedly as the speed reduces. 2. the stator windings being correctly connected in either star or delta. Phase currents become unbalanced, Plug braking windings generate increased heat, and torque is greatly Plug braking for an induction motor is simply a matter of reduced. Refer to section 12. l I. I for more details. reconnecting for the reverse direction of rotation while it is 3. the three line voltages being connected to the 1notor still rotating in a forward direction. While the d.c. motor windings. When any one supply line is not able to needed some form of current limiting, the induction supply current to the winding to which it is connected, motor does not since the current drawn is substantially the the condition known as "single-phasing" occurs. For same as the D.O.L. starting current. Once the motor has further details refer to section 12.1 1.2. stopped, some form of switching is needed to prevent it 4. the condition of.both windings in the 1noto1: The stator accelerating up to speed in the reverse direction. This is windings connected to the supply are pron1inent and often achieved by a friction driven contact on the shaft of obvious areas of concern. Noisy operation and reduced the motor. torque of a three-phase n1otor can mean that the bars of the rotor-the second winding-are in need of attention. Many cages consist of aluminium cast into shape in the laminations, and little can be done in the 12.11 Abnormal operating conditions way of niaintenance; many of the larger motors, The following applies specifically to three-phase motors. however, have prefabricated bars and rings of copper For further information on abnormal operating which are welded into place. It is possible to repair conditions applying to both single and three-phase dan1age to these ite1ns, whether they are broken or motors see section 13.6 in the next chapter. sin1ply loose in the rotor.

L, L, L, I I I I IE t-----+---+-r-., I

r;;\ l>y,1 I K2.4 ~ K2.1-2.3 - j I~ K1/5 K2/4 K3/1 + - I ~-____. Power circuit Control circuit

Fig. 12.36 Braking an a.c. motor by d.c. injection 242 ELECTRICAL PRINCIPLES FOR THE ELECTRICAL 1 c -8

c

A, A A B I I I Reversed I phase I s, I I (a) (b) Fig. 12.37 Reversal of one phase tor a star-connected motor

12.11.1 Phase reversal usually emits a "growling" noise and has an a Previously in this chapter the operation of the induction vibration due to the sustained high current valt motor has been based on the assumption that the motor had three identical windings and three equal currents 12.11.2 Single-phasing flowing in them, all spaced at 120° E to each other. If one Single-phasing is a condition that occurs when c phase is reversed, however, as shown in Figure l 2.37(a) a three-phase supply is open-circuited and is no for a star connection, these conditions no longer hold supply current to a three-phase load. The nar true. Two of the three currents that flow are at 60° E to used when one of the three phase windings in each other and the load system is unbalanced (Fig. open-circuited. 12.37(b)). The same condition applies to delta-connected The condition for single-phasing in a star-( loads, and both connections were discussed in Chapter 10. load is shown in Figure 12.38 (a), and it can be As a result of this incorrect connection, the motor a break in either the line or the phase windin. loses most ofits torque and is often unable to start against the circuit to a single current path. even a light load. If able to start at all, it usually rotates Figure 12.38(b) shows an open-circuited I very slowly and has unequal values of current in the phase delta-connected motor. There is one main cur windings. The values of current approach those drawn from L1 to L3 through phase A and another l during normal starting, but remain high. The motor L 1 to L3 through phases B and C in series. Bot!

I ~---- L, 0------, '1 -~---- .... I \ L,0-----~ \ A \ \ \ x I \ I B B I ' \ L2 o-x---~, ...... JL2 0---1--fY--VYl ______I L20-X_] \ - -, / ---- I ______I JL,()------J L, (_}------' ------·--/ / (a) (b) (c)

Fig. 12.38 Circuit conditions causing single-phasing with a three-phase motor THREE-PHASE INDUCTION MOTORS 243 are in parallel with each other, although not necessarily than normal currents in the parts of the circuit still in phase with each other. In Figure 12.38(c), a delta­ operating, with values approaching starting current connected induction motor is shown with phase C open­ values in some circumstances. It can also emit a low­ circuited. There are two current paths-L1 through phase pitched "growling" noise similar to that which occurs A to L3and L, through phase B to L,. during a phase reversal. Ifsingle-phasing occurs while the In each of the cases shown in Figure 12.38, the rotat­ motor is operating at normal speeds, the normal ing magnetic field is either destroyed or unbalanced, and humming sound often changes to a higher-pitched whine. causes unsatisfactory operation of the motor. The motor For any of the conditions for single-phasing outlined rotates at slower speeds, if it is able to start at all, because above, the VJ ratios between line and phase values are no of a much reduced starting torque. It usually draws higher longer valid.

Exercises 12.1 Briefly describe how the rotating magnetic 12.8 What is meant by: field is produced in a three-phase motor. (a) synchronous speed of an induction 12.2 (a) Define the term "synchronous speed''. motor? (b) Make a table showing the synchronous (b) actual speed? speeds of two-, four-, six- and eight-pole (c) slip speed? induction motors for frequencies of 40, 50 (d) What is the relationship between each and 60 Hz. of these three speeds? 12.3 Explain why an induction motor runs at less 12.9 Sketch a typical torque/speed curve for an than synchronous speed. induction motor having a normal squirrel­ cage rotor. At low values of slip, how does the 12.4 Explain why the power factor of an induction torque vary with load? What occurs when motor increases with the load. breakdown torque is reached? 12.5 Briefly describe the construction of the 12.10 Why do squirrel-cage motors take relatively squirrel-cage and the wound-rotor. large amounts of current when connected D.0.L.? 12.6 List three methods by which the starting 12.11 Discuss speed control by a method of current of a three-phase squirrel-cage in­ changing the number of poles in a motor. duction motor may be reduced. 12.12 Draw a circuit diagram for a star-delta starter 12.7 What is the disadvantage in starting a using push buttons for its initiation. Discuss squirrel-cage induction motor on a reduced the operation of the circuit and list two uses voltage? for this type of starter.

Problems 12.13 Determine the percentage slip for the (b) 1.5 kW, 940 r/min; following three-phase, 50 Hz motors: (c) 12 kW, 720 r/min; (a) four-pole, 1420 r/min; (d) 5 kW, 1450 r/min. (b) six-pole, 960 r/min; 12.17 On full load, a three-phase, 415 V, 50 Hz, (c) eight-pole, 720 r/min. six-pole motor draws 19 A at a power factor of 12.14 A 15 kW, three-phase, 415 V, 50 Hz, four-pole 0.85. If the torque developed is 95 Nm and the induction motor draws 190 A when started slip is 7%, calculate the efficiency of the D.O.L. in delta. Determine the starting motor. current using: 12.18 A cutting tool exerts a tangential force of 400 (a) the star-delta method; N on a 90 mm diameter steel bar which is (b) the autotransformer method (60% tap­ rotating at 145 r/min in a lathe. The efficiency ping). of the lathe gear train is 62% and the three­ phase, 415 V motor efficiency is 81%. 12.15 At full load the efficiency of the motor in Calculate the motor current if the power problem 12.14 is 83%, the power factor is 0.84 factor is 0.83. and the slip is 4%. Determine: 12.19 The rotor speed of a 10 kW, 415 V, three­ (a) the torque developed; phase, four-pole motor is 1455 r/min when it (b) the current drawn. operates from a source of 50 Hz. Find: 12.16 Calculate the full-load torque of each of the (a) synchronous speed; following motors: (b) slip speed; (a) 7.5 kW, 1440 r/min; (c) frequency of rotor currents.